MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 9791 CHEMISTRY. 9791/02 Paper 1 (Part A Written), maximum raw mark 100

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1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Pre-U Certificate MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 9791 CHEMISTRY 9791/02 Paper 1 (Part A Written), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, Pre-U, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

2 Page 2 Mark Scheme: Teachers version Syllabus Paper 1 (a) n(mg) = 9.0 g / 24.3 g mol 1 = 0.37 mol (1) [1] Allow two or more significant figures. (b) n(h 2 O) reacted = 2 n(mg) = 0.74 mol (1) Mass of water reacted = 0.74 mol 18 g mol 1 = 13.3 g Mass of excess water = 30 g 13.3 g = 16.7 g (1) [2] Allow two or more significant figures. (c) Vol of H 2 = 0.37 mol 24 dm 3 mol 1 = 8.9 dm 3 (1) [1] Allow two or more significant figures. (d) r H o = 924.5kJ mol 1 ( kJ mol 1 ) = kj mol 1 [2] 1 mark for correct signs; 1 mark for multiplying value for water by 2 Allow two or more significant figures. (e) Heat energy = kj mol mol= 131 kj (1) [1] Allow two or more significant figures. (f) Heat energy = (60 15) K 150 g 4.2 J g 1 K 1 = 28 kj (1) [1] Allow up to 4 significant figures. (g) The same amount of heat energy is released from the lumps (1) The rate of reaction (or the rate of heat generation) is slower and so a lower temperature will be reached (due to imperfect insulation) (1) / Allow temperature reached being the same if there is the stated assumption that the system is perfectly insulated. (1) VALID ALTERNATIVE: not all of the magnesium reacts as it becomes covered in insoluble magnesium hydroxide. (1) Therefore less energy released and lower temperature reached. (1) [2] (h) (i) CaO + H 2 O Ca(OH) 2 (1) 7 <ph 12 (1) [2] (ii) P 4 O H 2 O 4H 3 PO 4 OR P 2 O 5 + 3H 2 O 2H 3 PO 4 (1) 0 ph < 7 (1) [2] (iii) 6CaO + P 4 O 10 2Ca 3 (PO 4 ) 2 OR 3CaO + P 2 O 5 Ca 3 (PO 4 ) 2 (1) [1] Ignore state symbols. [Total: 15]

3 Page 3 Mark Scheme: Teachers version Syllabus Paper 2 a (i) Energy change to break one mole of bonds in the gas phase. 1 mark for each underlined point. [3] (ii) r H o = 2 ( ) kj mol 1 = 244kJ mol 1 1 mark for bonds broken; 1 mark for bonds made; 1 mark for correct sign if the answer is correct [3] (b) (i) Energy change = ( (2 4180))cm 1 = 11 cm 1 [1] (ii) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Superscripts must be used. [1] (iii) At least one K 4s atomic orbital labelled (1) Labelled sigma bond below labelled sigma antibond (1) (A single electron (spinning in either sense) in each atomic orbital and) two spin-paired electrons in the sigma bond (1) [3] Electrons must be shown with a single- or double-headed arrow. (iv) The outer electron in K is closer to the nucleus than the outer electron in Rb (1) There is less shielding of the nucleus for the K outer electron than the Rb outer electron. (1) (Despite the extra nuclear charge in rubidium) there is a weaker attraction of the electron to the nucleus (1) Allow the opposite statements with respect to Rb. [3] (v) Labelled Rb 5s orbital shown higher in energy than labelled K 4s orbital (1) Sigma bond is lower in energy than K 4s orbital and the antibond is higher in energy than the Rb 5s orbital (1) The bonding and antibonding orbitals must be labelled for the second mark. [2] (vi) E = 11 cm 1 h c N A 100 cm m 1 / 1000 J kj 1 = 0.13 kj mol 1 (1) [2] Two marks for correct answer. Deduct one mark for each error. One mark if final answer is out by a factor of N A i.e Allow two or more sig figs. [Total: 18]

4 Page 4 Mark Scheme: Teachers version Syllabus Paper 3 (a) (i) Point plotted corrected (must be within the correct small square in the grid) [1] (ii) Bonding is intermediate-covalent-ionic-metallic [1] (b) (i) NO 2 + 3e + 4H + ½ N 2 + 2H 2 O OR 2NO 2 + 6e + 8H + N 2 + 4H 2 O [2] 1 mark for correct number of electrons on the left hand side 1 mark for the rest of the balanced half equation (ignoring charge) (ii) 3CH 4 + 8NO 2 + 8H + 3CO 2 + 4N H 2 O [1] (iii) Enzyme catalysis [1] (c) (i) Oxidation state = {(2 112) 8} / 36 = (+)6 [1] (ii) [Mo 9 O 28 (H 2 O) 4 ] 2 OR [Mo 9 O 32 H 8 ] 2 [1] [Total: 8]

5 Page 5 Mark Scheme: Teachers version Syllabus Paper 4 (a) Carbon atom circled or otherwise indicated [1] (b) Nucleophile [1] Allow Nucleophilic or Lewis base or Lone-pair donor (c) Any unambiguous structure of the hemiaminal [1] OH H NH 2 No mark if atom connectivity isn t correct, e.g. OH CH 2 NH 2 (d) Addition [1] No credit for Electrophilic addition Allow Nucleophilic addition or Reduction. (e) Methanal (allow any carbonyl compound) [1] (f) Hydrolysis [1] Allow Hydration + Elimination but not Substitution + Elimination (g) Methanal: FGL 2 (1) After Reaction 2: FGL 2 (1) After Reaction 3: FGL 1 (1) [3] Accept equivalent names for the functional group levels. (h) (i) Allow any unambiguous structure for z. [1] O H (ii) 2-methylbutanal [1] Ignore incorrect use of spaces/hyphens but do not allow 2-methylbutan-1-al

6 Page 6 Mark Scheme: Teachers version Syllabus Paper (i) 1 mark for a correct structure NH 2 NH 2 H H 3 C O O H CH 3 OH OH 2nd mark for showing two optical isomers clearly with hashed and wedge bonds. [2] [Total: 13]

7 Page 7 Mark Scheme: Teachers version Syllabus Paper 5 (a) Mass of HCl = ¼ 55.6 mol 36.5 g mol 1 = 507 g [1] No sig figs or units penalties. (b) Amount of NaOH = dm mol dm 3 = mol (1) Amount of HCl in volumetric flask = mol = mol (1) [HCl] = mol / dm 3 = 12.4 mol dm 3 (1) Final answer to 3 sig figs (1) [4] (c) (i) H 2 SO 4 + NaCl HCl + NaHSO 4 [1] Ignore state symbols. Allow H 2 SO 4 + 2NaCl 2HCl + Na 2 SO 4 (ii) H 2 SO 4 + 2HBr Br 2 + SO 2 + 2H 2 O [1] Ignore state symbols Allow H 2 SO 4 + 2HBr Br 2 + H 2 SO 3 + H 2 O (iii) Sulfuric acid is the oxidising agent (1) No credit for S being the oxidising agent. The oxidation number of bromine increases (from 1 to 0) OR the oxidation number of sulfur decreases (from +6 to +4) (1) [2] (d) (i) Bond strength decreases because the bonds gets longer OR because there is greater shielding of the bonding electrons from the halogen nucleus due to the additional inner shells of electrons. [1] No credit for answers based on electronegativity or ionic radii. (ii) Acidic strength increases because the H Hal bond gets weaker. [1] (iii) Increasing boiling point for HCl HBr HI due to increasing van der Waals (permanent) dipole (permanent) dipole forces. (1) HF boiling point higher than HCl due to hydrogen bonding. (1) [2] [Total: 13]

8 Page 8 Mark Scheme: Teachers version Syllabus Paper 6 (a) Molecular formula = C 4 H 4 O 2 [1] (b) Correct structure (1) O OH Name = but-2-ynoic acid (1) [2] No mark for name if it is inconsistent with the structure given. (c) %C = (24/42) 100% = 57.1% %H = (2/42) 100% = 4.8% %O = (16/42) 100% = 38.1% [2] 2 marks all correct. 1 mark for two out of three correct. Don t penalise two or more significant figures. Allow 5% for H. (d) m/z = 84 [1] (e) (i) Allow wavenumber values from the following ranges Double bonds: cm 1 (inclusive) Single bonds (no H): 1500 cm 1 Single bonds to H: cm 1 (inclusive) Allow a correct range rather than a value. 2 marks for 3 correct responses 1 mark for 1 or 2 correct responses [2] (ii) Broad OR a wavenumber range given that is at least 500 cm 1 across. [1] (f) O OH [1] [Total: 10]

9 Page 9 Mark Scheme: Teachers version Syllabus Paper 7 (a) 1 mark for each correct test; 1 mark for each correct observation; 1 mark for each correct deduction. Halogenoalkanes: max 5 marks Testing for the haloalkanes with aqueous silver nitrate or aqueous lead nitrate (1) White ppt (produced slowly) (1) with the chloroalkane (1) Yellow ppt (produced quickly) (1) with the iodoalkane (1) Allow use of NaOH to hydrolyse the halogenoalkane first (though the alkali precipitates out the silver). [5] Aldehyde: max 3 marks Tollens (or Fehling s etc) reagent or ammoniacal silver nitrate (1) used to identify the aldehyde (1) Silver mirror (or red ppt) produced (1) Brady s reagent or 2,4-dinitrophenylhydrazine or 2,4-DNP (1) used to identify the aldehyde (1) Orange ppt (1) Ignore oxidation of the aldehyde if it has already been identified. [3] Alcohols: max 6 marks Identification of alcohols by their oxidation [max 3 marks] Oxidising agent specified to identify the alcohols (dichromate/ manganate must be acidified) (1) Colour change specified (1) Identification of the two alcohols as having given the colour change. (1) Distinguishing between the two alcohols using their oxidation [max 3 marks] Test to distinguish the two alcohols, eg distilling off oxidation product (1) Product of oxidising propan-2-ol distils off readily / is not acidic (1) Product of oxidising propan-1-ol distils off at higher temperature / is acidic (1) Lucas reagent method: Lucas reagent or conc HCl with ZnCl 2 catalyst (1) Very slow cloudiness (1) implies propan-1-ol (1) Moderately rapid cloudiness (1) implies propan-2-ol (1) Alternative methods Sodium (1) specified to identify the alcohols (1) Effervescence (or hydrogen gas produced) (1) Glacial ethanoic acid and conc sulfuric acid (1) to identify the alcohols (1) Sweet / fruity vapour produced (1) Iodoform test (1) to identify propan-2-ol (1) Iodoform observations for positive test (1) If alcohols are identified by a concise clear method, e.g. Lucas reagent, which yields only 5 marks, additional mark to be given for economy of method (1) [6] Alkane: max 1 mark Hexane gives no positive results with any of these tests [1] Give credit to alternative legitimate methods and also to the observations from those methods [Max. 15 marks]

10 Page 10 Mark Scheme: Teachers version Syllabus Paper (b) Reduction (max 5 marks) Dissolve benzophenone in ethanol (if using NaBH 4 ) or dry ether (if using LiAlH 4 ) (1) Choice of NaBH 4 or LiAlH 4 as reducing agent (1) Dissolve reducing agent in ethanol or water (if using NaBH 4 ) or dry ether (if using LiAlH 4 ) (1) Allow ether as a solvent for NaBH 4. Use excess reducing agent (1) Heat (under reflux) benzophenone with reducing agent (1) Separation (max 3 marks) If using alcohol: Precipitate product (1) by adding (excess) water (1) Then filter off product (1) If using ether: Add (excess) water (1) Separate the ether layer (with a separating funnel) (1) Then distil off ether to recover product (1) Allow Alternative Separation Method (based on solubility in warm water) (max 3 marks) Add water and warm mixture (1) Filter off undissolved X (allow decant) (1) Evaporate off water or leave to cool to recover Y (1) Marks for separation reliant on a feasible technique, ie a reasonable sequence of steps that would work. Purification (max 3 marks) Wash product with cold water or cold hexane (1) Recrystallise product from (a minimum of) hot hexane (or water) (1) Cooling/scratching of glass to aid precipitation (1) Checking Purity (max 1 mark) Check purity of product by measuring its melting point OR by thin-layer chromatography (against starting material) OR infrared (1) [Max: 8 marks] [Total: 23]

Paper 2 Part A Written For Examination from 2016 SPECIMEN MARK SCHEME 2 hours 15 minutes MAXIMUM MARK: 100

Cambridge International Examinations Cambridge Pre-U Certificate www.xtremepapers.com CHEMISTRY (PRINCIPAL) 9791/02 Paper 2 Part A Written For Examination from 2016 SPECIMEN MARK SCHEME 2 hours 15 minutes